Notice the jump in this argument. We are given nothing related to Z in our premise.
We only know Y-some-X so we need to incorporate Z to properly arrive at the conclusion Y-some-Z.
This means we need X to be sufficient for Z as follows:
Y-some-X ==> Z
So our missing premise is:
X ==> Z not Z ==> not X
This allows us to properly conclude: Y-some-Z (which can be reversed to Z-some-Y)
Hope that helps! For a more in-depth discussion of these concepts, please watch our video lesson on Quantifiers.
elviaeolveraSeptember 1, 2017
Does that mean P: "some x are y" "some y are x"
Is that how it is read? I didn't run into any of these in my LSAT review
MehranSeptember 3, 2017
Yes, that is correct. Remember that "some" means "at least 1." If some Xs are Ys, then that means at least one Y is an X, i.e., that some Ys are Xs.
Hope this helps! Please let us know if you have any additional questions.
bball7January 13, 2019
I imagine it as multiplying an inequality by -1. If we know that if X some Y, we can further conclude that if X then Z (every X also means a Z) then we know that Z some Y. Now since X => Z isn't an option, we look for the opposite which is notZ = notX. The direction of the arrow flips when you convert the nots. Is this a correct way to look at it?
